Perl regex to change date format

I'm totally new to Perl, and I need to get a small find and replace done to change the date format in a set of l开发者_StackOverflow社区arge file. The files have dates in the format: dd.mm.yyyy and I need to change them to: mm-dd-yyyy

How do I do this with perl?

I have a basic code to read through the files in a directory and write to an output file, I'll need to have my replace logic in-between the while loop (if i'm not wrong!).

#!c:/perl64/bin/perl.exe

#loop around a directory
@files = <C:/perl64/data/*>;

# loop around files
foreach $file (@files) {

#Read File
open READ, $file or die "Cannot open $read for read :$!";

#Output File
$fname=substr($file, rindex($file,"/")+1,length($file)-rindex($file,"/")-1);
$write="C:/perl64/output/$fname";

open WRITE, ">$write" or die "Cannot open $write for write :$!";

#Loop Around file
while (<READ>) {

           # TO DO: Change date format from dd.mm.yyyy to mm-dd-yyyy

           #Write to ourput file
           print WRITE "$_";

}

} 

Regards,

Anand

You can use the substitution operator s///:

while (<READ>) {
    s/(\d{2})\.(\d{2})\.(\d{4})/$2-$1-$3/;
    print WRITE "$_";
}

Here's a simple script that will take optional arguments for input and output directories.

The use of opendir instead of a glob will save you some trouble cleaning up the file names.

use strict;
use warnings;
use autodie;

my $indir = shift || "C:/perl64/data";
my $outdir = shift || "C:/perl64/output";

opendir(my $in, $indir);

while (readdir $in) {
    next unless -f;
    open my $infile, '<', $_;
    open my $outfile, '>', $outdir . "/" . $_;
    while (<$infile>) {
        s/([0-9]{2})\.([0-9]{2})\.([0-9]{4})/$2-$1-$3/g; 
        print $outfile $_;
    }
}